Chemistry II
Midterm Exam
18 April, 2014 (Total Score: 104 points)
Periodic Table of Elements
Constants
R = 8.314 J/mol·K = 0.08314 L·bar/K·mol = 0.0821 L·atm/K·mol = 8.314 L·kPa/K·mol
1 bar = 750.06 torr = 0.9869 atm
F = 9.6485×104 C/mol
1.
Professor X created two synthetic chemicals, A and B, which both are nonelectrolytes. Professor
X asked one of his student, Dudu, to find out which compound has the higher standard enthalpy
of vaporization. Dudu went back to measure their vapor pressure, and the results are listed
below:
 The vapor pressure of A is 517.5 Torr at 70°C and 58.24 Torr at 25°C.
 The vapor pressure of B is 400.0 Torr at 70°C and 96.44 Torr at 25°C.
Something interesting bothers Dudu. At 25°C, the vapor pressure of A is lower than B, but at
70°C, the vapor pressure of A is higher than B. Dudu cannot judge which compound has the
higher standard enthalpy of vaporization. Now, it is your job to use the proper thermodynamic
equation and help Dudu to solve the puzzle.
1
(a) Among the following list, which thermodynamic equation can help you to solve the
puzzle? (2%)
(i) The Nernst Eqution.
(ii) The van’t Hoff Equation
(iii) The Clausius-Claypeyron Equation
(iv) The Gibbs-Duhem Equation
(v) The Raoult’s law
(b) Which compound has the higher standard enthalpy of vaporization? Please calculate the
standard enthalpy of vaporization for both A and B to support your answer. (8%)
(c) Which compound has the more negative standard Gibbs free energy of vaporization at
25°C? Please explain it briefly. (4%)
(d) Which compound has higher normal boiling point? Please calculate the normal boiling
point for both A and B to support your answer. (8%)
(e) At 25°C, Dudu mixed A and B together. Professor X found that a liquid mixture of A and B
is an ideal solution. But, Dudu did not know what an ideal solution is. Please answer the
following questions to help Dudu to understand the ideal solution.
(i) The standard enthalpy of the mixing does not change. True or false? (2%)
(ii) The standard Gibbs free energy of the mixing does not change. True or false? (2%)
(iii) The liquid mixture must follow Raoult’s law. True or false? (2%)
(iv) The liquid mixture may have the azeotrope. True or false? (2%)
2.
A solid consists of a mixture of NaNO3 and Mg(NO3)2. When 6.5 g of this solid is dissolved in
50.0g of water, the freezing point is lowered down by 5.50°C. What is the composition of the
solid (by mass)? kf of water is 1.86 K/m. (8%)
3.
(a) Calculate the reaction Gibbs free energy of 2 I(g) → I2(g) at 1200 K (K = 0.147) when the
partial pressures of I and I2 are 0.98 bar and 0.13 bar, respectively. (b) What is the spontaneous
direction of the reaction? Explain briefly. (total 6%)
4.
A reaction used in the production of gaseous fuels from coal, which is mainly carbon, is
C(s) + H2O(g) ⇌ CO(g) + H2(g). (a) Evaluate Kc at 900 K, given that the standard Gibbs free
energies of formation of CO(g) and H2O(g) at 900 K are -191.28 kJ/mol and -198.08 kJ/mol,
respectively. (b) A 5.20-kg sample of graphite and 125 g of water were placed into a 10.0-L
container and heated to 900 K. What are the equilibrium concentrations?
(total 10%)
5.
Calculate the equilibrium constant at 25°C and 150°C for the reaction
NH4Cl(s) ⇌ NH3(g) +
HCl(g) with data given. H°f (in kJ/mol): NH3(g) = -46.11; HCl(g) = -92.31; NH4Cl(s) = -314.43.
S° (in J/K·mol): NH3(g) = 192.45; HCl(g) = 186.91; NH4Cl(s) = 94.60. (8%)
2
6.
Predict the response of the equilibria in the Haber process and explain briefly:
(a) Compress the mixture. (2%)
(b) Heat the mixture. (2%)
(c) Add catalyst. (2%)
Select correct answer for question 7 to 15. (2 % for each question)
7. Which of the following sets of conditions must be maintained when determining the Eo of an
electrochemical cell?
(a) 0 oC, 0.1 M solution, 100 kPa
(b) 25 oC, 0.1 M solution, 100 kPa
(c) 25 oC, 1.0 M solution, 101.3 kPa (d) 0 oC, 1.0 M solution, 101.3 kPa
8.
Electrons in an electrochemical cell move
(a) from the oxidizing agent to the reducing agent by means of a wire
(b) from anode to cathode by means of a salt bridge
(c) from anode to cathode by means of a wire
(d) from cathode to anode by means of a wire
9.
A particular reaction has a predicted Eo value of 1.35 V. This indicates that the reaction would
(a) react rapidly (b) consume energy (c) go to completion (d) be spontaneous
10. What is the unit for the rate of charge flow?
(a) Volt (b) Joule (c) Ampere (d) Coulomb (e) Ohm
11. Which of the following energy sources will most likely have a PbO2 electrode?
(a) dry battery (b) automotive battery (c) Li-ion battery (d) mercury battery
12. Which one of the following converts chemical energy to electrical energy?
(a) industrial production of aluminum (b) copper plating by electrolysis
(c) discharging of a Li-ion battery (d) industrial production of chlorine
13. Which one of the following reactions is NOT a redox reaction?
(a) H2CO3(aq) → H2O(l) + CO2(g) (b) 2 HgO(s) → 2 Hg(l) + O2(g)
(c) H2(g) + Br2(g) → 2 HBr(g)
(d) 2 HCl(aq) + Zn(s) →H2(g) + ZnCl2(aq)
(e) 2 KClO3(s) → 2KCl(s) + 3 O2(g)
14. Which of the following is the best oxidizing agent?
(a) F2(g) (b) Ag+(aq) (c) Ag(s) (d) Al3+(aq) (e) K(s)
3
15. Which one of the following will decrease the reduction potential for the half cell:
MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(l)
(a) add H2SO4(l) (b) decrease [Mn2+(aq)] (c) increase [MnO4–(aq)] (d) add NaOH(s)
16. For the following redox equation which occurs in an acidic solution, answer the following
questions.
N2H4(g) + BrO3–(aq) → Br2 (l) + N2(g)
(a) What is the oxidation half-reaction? (2%)
(b) What is the reduction half-reaction? (2%)
(c) Balanced the redox equation. (4%)
(d) By assuming that a salt bridge and platinum electrodes are used to construct a cell, write a
cell notation. (2%)
(e) At the standard state, the oxidation half-reaction has a potential 1.21 V while the reduction
half-reaction has a potential 1.531 V. Estimate the Gr° and the cell potential.
(8%)
4
102B Chemistry (II) midterm exam
Answer
1.
(a) (iii) The Clausius-Claypeyron Equation (2%)
(b) (Total 8%) The compound A has the higher standard enthalpy of vaporization. (2%)
First, the standard enthalpy of vaporization for A could be obtained
°
−∆𝐻𝑉𝑎𝑝
58.24
1
1
ln (
)=
(
−
)
517.5
8.314 273 + 25 273 + 70
J
kJ
°
∆𝐻𝑉𝑎𝑝
= 41252.23 mol = 41.25 mol (3%)
Second, the standard enthalpy of vaporization for B could be obtained
°
−∆𝐻𝑉𝑎𝑝
96.44
1
1
ln (
)=
(
−
)
400.0
8.314 273 + 25 273 + 70
J
kJ
°
∆𝐻𝑉𝑎𝑝
= 26864.17 mol = 26.86 mol (3%)
(c) (Total 4%) The compound B has the more negative standard Gibbs free energy of vaporization
at 25°C. (2%)
The reason is that the vapor pressure of B is larger than A at 25°C. (2%)
(d) (Total 8%) The compound B has higher normal boiling point. (2%)
The normal boiling point of A is the temperature at which the vapor pressure of A is 760 torr.
760
−41252.23 1
1
ln (
)=
(
−
)
58.24
8.314
𝑇𝑛𝑏 298
𝑇𝑛𝑏 = 352.36 𝐾 = 79.36 ℃ (3%)
The normal boiling point of B is the temperature at which the vapor pressure of B is 760 torr.
760
−26864.17 1
1
ln (
)=
(
−
)
96.44
8.314
𝑇𝑛𝑏 298
𝑇𝑛𝑏 = 368.08 𝐾 = 95.08 ℃ (3%)
(e) (Total 8%) (i) True (2%)
(ii) False (2%)
2.
(iii) True (2%)
(iv) False (2%)
using the freezing point depression constant:
∆𝑇 = 𝑖𝑘𝑓 𝑚
5.50 = i × 1.86 × (molality)
i × (molality) = 2.957
find the moles of ions in with 50 grams of water:
50 𝑔
2.957 × 1000 𝑔 = 0.1478 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑖𝑜𝑛𝑠
(2%)
5
Assume that the mass of NaNO3 is X g, and then the mass of Mg(NO3)2 is (6.5-X) g.
Here, each mole of NaNO3 releases 2 moles of ions, and each mole of Mg(NO3)2 releases 3
moles of ions.
First, we calculate the moles of ions released from dissolving NaNO3
X g NaNO3
g
NaNO3
mol
84.99
× 2 = 0.02353 X moles of ions from NaNO3 (2%)
Second,, we calculate the moles of ions released from dissolving Mg(NO3)2
(6.50 − X) g Mg(NO3 )2
g
Mg(NO3 )2
mol
148.32
× 3 = ( 0.1315 − 0.02023 X) moles of ions from Mg(NO3 )2
(2%)
Now, we add up the total moles of ions:
(0.02353 X) + (0.1315 - 0.02023 X ) = 0.1478 moles of ions found by freezing point depression
0.0033 X = 0.0163
X = 4.94 grams of NaNO3
which leaves 1.56 g Mg(NO3)2 (2%)
正確答案也可以使用重量百分比：
4.94 g NaNO3 / 6.50 g total = 76.0 % NaNO3 ; which leaves 24.0% Mg(NO3)2
(2%)
如果答案正確，算式不太一樣也算全對。可得 8 分。
3.
The free energy at a specific set of conditions is given by
Gr = Gr + RT ln Q = - RT ln K + RT ln Q = - RT ln K + RT ln([I2]/[I]2)
= -(8.314 J/K mol)(1200 K) ln(0.147) + (8.314 J/K mol)(1200 K) ln[(0.13)/(0.98)2]
= - 0.83 kJ/mol
Because Gr is negative, the reaction will be spontaneous to produce I2.
4.
注意，Exercise 10.83 課本解答有誤，將 K 當作 Kc
Gro
RT
Gro = (-191.28 kJ · mol-1) – (-198.08 kJ · mol-1) = 6.8 kJ · mol-1
(a) Gro =  RT lnK;
lnK = 
6800 J  mol 1
= -0.909;
(8.314 J  K 1  mol 1 )(900K )
K = 0.403 ; Kc = K(RT)-1 = 0.403·(0.08314·900)-1 = 5.39×10-3
1mol C
(b) 5.20 103 g C 
= 433. mol C
12.011g C
lnK = 
6
125 g H2O 
1mol H 2O
=
18.016 g H 2O
6.94 mol H2O , H2O is limiting.
Conc. of H2O = 6.94 mol/10.0 L = 0.694 mol·L-1
Concentration (mol/L) H2O (g)
CO(g)
initial
0.694
0
change
x
x
final
0.694-x
x
KC =
COH 2 
H 2O
=
H2(g)
0
x
x
( x)( x)
= 5.39×10-3
(0.694  x)
x2 = 3.74×10-3 – (5.39×10-3)x
x = 0.0585 (mol/L)
[CO] = [H2] = 5.85×10-2 mol/L ;
5.
[H2O] = 6.36×10-1 mol/L
H°r = (-46.11) + (-92.31) - (-314.43) = +176.01 (kJ/mol)
S°r = (192.45) + (186.91) – (94.60) = 284.76 (J/K·mol)
At 298 K:
G°r = (+176.01×103) - 298×284.76 = 91.15×103 (J/mol)
K = exp(-91.15×103 /(298×8.314)) = 1.05×10-16
At 423 K:
G°r = (+176.01×103) - 423×284.76 = 55.56×103 (J/mol)
K = exp(-55.56×103 /(423×8.314)) = 1.38×10-7
6.
(a) n = -2, favor product, NH3.
(b) it is exothermic, favor reactants
(c) no effect on equilibrium, but increase the rate to equilibrium.
7. (c)
8. (c)
9. (d)
10. (c)
11. (b)
12. (c)
13. (a)
14. (a)
15. (d)
16. (a) oxidation half-reaction: N2H4(g) → N2(g) + 4 H+(aq) + 4 e– (2%)
(b) reduction half-reaction: 2 BrO3–(aq) + 12 H+(aq) + 10 e– → Br2(aq) + 6 H2O (2%)
(c) Balanced the redox equation:
oxidation half-reaction ×5 5 N2H4(g) → 5 N2(g) + 20 H+(aq) + 20 e– (1%)
reduction half-reaction ×2 4 BrO3–(aq) + 24 H+(aq) + 20 e– → 2 Br2(aq) + 12 H2O (1%)
7
5 N2H4(g) + 4 BrO3–(aq) + 4 H+(aq) → 2 Br2(l) + 5 N2(g) + 12 H2O (2%)
(d) Pt(s)| N2H4(g) , N2(g)|H+(aq) || H+(aq) | Br2(l) , BrO3–(aq)|Pt(s) (2%)
(e) oxidation half-reaction: N2H4(g) → N2(g) + 4 H+(aq) + 4 e–
E1o = 1.21 V
reduction half-reaction: 2 BrO3–(aq) + 12 H+(aq) + 10 e– → Br2(aq) + 6 H2O
G1o (oxidation) = -n1FE1o
n1 = 4
G2o (reduction) = -n2FE2o
n2 = 10
Gro = -n3FE3o
E2o = 1.531 V
n3 = 20
Gro = 5×G1o (oxidation) + 2×G2o (reduction) = 5.289×106 J = 5289 kJ
(4%)
E3o = Gro/(-n3F) = 2.74 V 或 E3o = E1o + E2o = 2.74 V (4%)
8